Question

Pure acid is to be added to a 10% acid solution to obtain 90L of 81% solution. What amounts of each should be used?How many liters of 100% pure acid should be used to make the solution? 04

Answer 1

Let's use the variable x to represent the amount of pure acid and y to represent the amount of 10% acid.

Since the total amount wanted is 90 L, we can write the equation:

`x+y=90`

Also, the final solution is 81%, so we can write our second equation:

`100\cdot x+10\cdot y=81\cdot(x+y)`

From the first equation, we can solve for y and we will have y = 90 - x.

Using this value in the second equation, we have:

`100x+10(90-x)=81(x+90-x)`

Solving for x, we have:

`\begin{gathered} 100x+900-10x=81\cdot90 \\ 90x+900=7290 \\ 90x=7290-900 \\ 90x=6390 \\ x=(6390)/(90) \\ x=71 \end{gathered}`

Therefore the amount of pure acid to be used is 71 L and the amount of 10% acid is 19 L.